Scalars, Vectors, Tensors and All That

What Are Quaternions?
So What Is the Difference?

According to my math dictionary, a tensor is ...

An abstract object having a definitely specified system of components in every coordinate system under consideration and such that, under transformation of coordinates, the components of the object undergoes a transformation of a certain nature.

To make this introduction less abstract, I will confine the discussion to the simplest tensors under rotational transformations.  A rank-0 tensor is known as a scalar.  It does not change at all under a rotation.  It contains exactly one number, never more or less.  There is a zero index for a scalar.  A rank-1 tensor is a vector.  A vector does change under rotation.  Vectors have one index which can run from 1 to the number of dimensions of the field, so there is no way to know a priori how many numbers (or operators, or ...) are in a vector.  n-rank tensors have n indices. The number of numbers needed is the number of dimensions in the vector space raised by the rank.  Symmetry can often simplify the number of numbers actually needed to describe a tensor.

There are a variety of important spin-offs of a standard vector.  Dual vectors, when multiplied by its corresponding vector, generate a real number, by systematically multiplying each component from the dual vector and the vector together and summing the total.  If the space a vector lives in is shrunk, a contravariant vector shrinks, but a covariant vector gets larger.  A tangent vector is, well, tangent to a vector function.

Physics equations involve tensors of the same rank.  There are scalar equations, polar vector equations, axial vector equations, and equations for higher rank tensors.  Since the same rank tensors are on both sides, the identity is preserved under a rotational transformation.  One could decide to arbitrarily combine tensor equations of different rank, and they would still be valid under the transformation.

There are ways to switch ranks.  If there are two vectors and one wants a result that is a scalar, that requires the intervention of a metric to broker the transaction.  This process in known as an inner tensor product or a contraction.  The vectors in question must have the same number of dimensions.  The metric defines how to form a scalar as the indices are examined one-by-one.  Metrics in math can be anything, but nature imposes constraints on which ones are important in physics.  An aside: mathematicians require the distance is non-negative, but physicists do not.  I will be using the physics notion of a metric.  In looking at events in spacetime (a 4-dimensional vector), the axioms of special relativity require the Minkowski metric, which is a 4x4 real matrix which has down the diagonal 1, -1, -1, -1 and zeros elsewhere.  Some people prefer the signs to be flipped, but to be consistent with everything else on this site, I choose this convention.  Another popular choice is the Euclidean metric, which is the same as an identity matrix.  The result of general relativity for a spherically symmetric, non-rotating mass is the Schwarzschild metric, which has "non-one" terms down the diagonal, zeros elsewhere, and becomes the Minkowski metric in the limit of the mass going to zero or the radius going to infinity.

An outer tensor product is a way to increase the rank of tensors.  The tensor product of two vectors will be a 2-rank tensor.  A vector can be viewed as the tensor product of a set of basis vectors.

What Are Quaternions?

Quaternions could be viewed as the outer tensor product of a scalar and a 3-vector.  Under rotation for an event in spacetime represented by a quaternion, time is unchanged, but the 3-vector for space would be rotated.  The treatment of scalars is the same as above, but the notion of vectors is far more restrictive, as restrictive as the notion of scalars. Quaternions can only handle 3-vectors.  To those familiar to playing with higher dimensions, this may appear too restrictive to be of interest.  Yet physics on both the quantum and cosmological scales is confined to 3-spatial dimensions.  Note that the infinite Hilbert spaces in quantum mechanics a function of the principle quantum number n, not the spatial dimensions.  An infinite collection of quaternions of the form (En, Pn) could represent a quantum state.  The Hilbert space is formed using the Euclidean product (q* q').

A dual quaternion is formed by taking the conjugate, because q* q = (t^2 + X.X, 0).  A tangent quaternion is created by having an operator act on a quaternion-valued function

(d/dt, Del) acting on (f(q), F(q)) = (f dot - div F, F dot + Grad f + Curl F)

What would happen to these five terms if space were shrunk?  The 3-vector F would get shrunk, as would the divisors in the Del operator, making functions acted on by Del get larger.  The scalar terms are completely unaffected by shrinking space, because df/dt has nothing to shrink, and the Del and F cancel each other.  The time derivative of the 3-vector is a contravariant vector, because F would get smaller.  The gradient of the scalar field is a covariant vector, because of the work of the Del operator in the divisor makes it larger.  The curl at first glance might appear as a draw, but it is a covariant vector capacity because of the right-angle nature of the cross product.  Note that if time where to shrink exactly as much as space, nothing in the tangent quaternion would change.

A quaternion equation must generate the same collection of tensors on both sides.  Consider the product of two events, q and q':

(t, X) times (t prime , X prime) =  (t t prime  - X dot X prime , t X prime  + X t prime + X cross X prime)axial vectors: X cross X prime

Where is the axial vector for the left hand side?  It is imbedded in the multiplication operation, honest :-)

(t prime , X prime) times (t, X) = (t prime t - X prime  dot X, t prime  X + X prime  t + X prime  cross X) = (t t prime  - X dot X prime , t X prime  + X t prime  - X cross X prime)

The axial vector is the one that flips signs if the order is reversed.

Terms can continue to get more complicated.  In a quaternion triple product, there will be terms of the form (XxX').X".  This is called a pseudo-scalar, because it does not change under a rotation, but it will change signs under a reflection, due to the cross product.  You can convince yourself of this by noting that the cross product involves the sine of an angle and the dot product involves the cosine of an angle. Neither of these will change under a rotation, and an even function times an odd function is odd.  If the order of quaternion triple product is changed, this scalar will change signs for at each step in the permutation.

It has been my experience that any tensor in physics can be expressed using quaternions.  Sometimes it takes a bit of effort, but it can be done.

Individual parts can be isolated if one chooses.  Combinations of conjugation operators which flip the sign of a vector, and symmetric and antisymmetric products can isolate any particular term.  Here are all the terms of the example from above

(t,X) times (t prime ,X prime) = (t t prime  - X dot X prime , t X prime  + X t prime + X cross X prime)

scalars: t = (q + q conjugated) over 2, t prime  = (q prime + q prime  conjugated) over 2, t t prime  - X dot X prime  = (q q prime  + (q q prime) conjugated)/2

polar vectors: X = (q - q conjugated) over 2, X prime  = (q prime  - q prime  conjugated) over 2,  t X prime  + X t prime  = ((q q prime +(q prime  q)) - (qq prime +(q prime q)) conjugated) over 4

axial vectors: X cross X prime  = (q q prime -(q prime  q)) over 2

The metric for quaternions is imbedded in Hamilton's rule for the field.

i squared = j squared = k squared = i j k = - 1

This looks like a way to generate scalars from vectors, but it is more than that.  It also says implicitly that i j = k, j k = i, and i, j, k must have inverses.  This is an important observation, because it means that inner and outer tensor products can occur in the same operation. When two quaternions are multiplied together, a new scalar (inner tensor product) and vector (outer tensor product) are formed.

How can the metric be generalized for arbitrary transformations?  The traditional approach would involve playing with Hamilton's rules for the field.  I think that would be a mistake, since that rule involves the fundamental definition of a quaternion.  Change the rule of what a quaternion is in one context and it will not be possible to compare it to a quaternion in another context.  Instead, consider an arbitrary transformation T  which takes q into q'

q transform to q prime  = T times q

T is also a quaternion, in fact it is equal to q' q^-1.  This is guaranteed to work locally, within neighborhoods of q and q'.  There is no promise that it will work globally, that one T will work for any q.  Under certain circumstances, T will work for any q.  The important thing to know is that a transformation T necessarily exists because quaternions are a field. The two most important theories in physics, general relativity and the standard model, involve local transformations (but the technical definition of local transformation is different than the idea presented here because it involves groups).

This quaternion definition of a transformation creates an interesting relationship between the Minkowski and Euclidean metrics.

Let T = I, the identity matrix

(I q I q + (I q I q) conjugated) over 2 = (t squared - X dot X, 0), the Minkowski interval

(I q) conjugated times I q = (t squared + X dot X, 0), the Euclidean norm

In order to change from wrist watch time (the interval in spacetime) to the norm of a Hilbert space does not require any change in the transformation quaternion, only a change in the multiplication step. Therefore a transformation which generates the Schwarzschild interval of general relativity should be easily portable to a Hilbert space, and that might be the start of a quantum theory of gravity.

So What Is the Difference?

I think it is subtle but significant.  It goes back to something I learned in a graduate level class on the foundations of calculus.  To make calculus rigorous requires that it is defined over a mathematical field. Physicists do this be saying that the scalars, vectors and tensors they work with are defined over the field of real or complex numbers.

What are the numbers used by nature?  There are events, which consist of the scalar time and the 3-vector of space.  There is mass, which is defined by the scalar energy and the 3-vector of momentum.  There is the electromagnetic potential, which has a scalar field phi and a 3-vector potential A.

To do calculus with only information contained in events requires that a scalar and a 3-vector form a field.  According to a theorem by Frobenius on finite dimensional fields, the only fields that fit are isomorphic to the quaternions (isomorphic is a sophisticated notion of equality, whose subtleties are appreciated only by people with a deep understanding of mathematics).  To do calculus with a mass or an electromagnetic potential has an identical requirement and an identical solution.  This is the logical foundation for doing physics with quaternions.

Can physics be done without quaternions?  Of course it can!  Events can be defined over the field of real numbers, and then the Minkowski metric and the Lorentz group can be deployed to get every result ever confirmed by experiment.  Quantum mechanics can be defined using a Hilbert space defined over the field of complex numbers and return with every result measured to date.

Doing physics with quaternions is unnecessary, unless physics runs into a compatibility issue.  Constraining general relativity and quantum mechanics to work within the same topological algebraic field may be the way to unite these two separately successful areas.

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